Hello, Could you help me solve this ?

I'd appreciate ur help, thanks a lot

\(\displaystyle \frac{\log_2{(x^2 + 2x - 7)}}{\log_2{(x - 3)}} = 1\)

\(\displaystyle \log_2{(x^2 + 2x - 7)} = \log_2{(x - 3)}\)

\(\displaystyle x^2 + 2x - 7 = x - 3\)

\(\displaystyle x^2 + x - 4 = 0\)

\(\displaystyle x^2 + x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 - 4 = 0\)

\(\displaystyle \left(x + \frac{1}{2}\right)^2 - \frac{17}{4} = 0\)

\(\displaystyle \left(x + \frac{1}{2}\right)^2 = \frac{17}{4}\)

\(\displaystyle x + \frac{1}{2}= \pm \frac{\sqrt{17}}{2}\)

\(\displaystyle x = -\frac{1}{2} \pm \frac{\sqrt{17}}{2}\).

Therefore

\(\displaystyle x = \frac{-1 - \sqrt{17}}{2}\) or \(\displaystyle x = \frac{-1 + \sqrt{17}}{2}\).